在numpy数组中迭代没有for循环 [英] Iterating without for loop in numpy array

问题描述

我需要对numpy数组进行逻辑迭代，其值取决于其他数组的元素。我在下面写了代码来澄清我的问题。有没有for循环解决这个问题的任何建议？

I need to do logical iteration over numpy array, which's values depend on elements of other array. I've written code below for clarifying my problem. Any suggestions to solve this problem without for loop?

Code
a = np.array(['a', 'b', 'a', 'a', 'b', 'a'])
b = np.array([150, 154, 147, 126, 148, 125])
c = np.zeros_like(b)
c[0] = 150
for i in range(1, c.size):
    if a[i] == "b":
        c[i] = c[i-1]
    else:
        c[i] = b[i]

推荐答案

这是一种使用 np.maximum.accumulate 和 np.where 创建将成为的步进索引按一定的时间间隔停止，然后只需索引到 b 即可获得所需的输出。

Here's an approach using a combination of np.maximum.accumulate and np.where to create stepped indices that are to be stopped at certain intervals and then simply indexing into b would give us the desired output.

因此，实现将是 -

Thus, an implementation would be -

mask = a!="b"
idx = np.maximum.accumulate(np.where(mask,np.arange(mask.size),0))
out = b[idx]

逐步示例运行 -

In [656]: # Inputs 
     ...: a = np.array(['a', 'b', 'a', 'a', 'b', 'a'])
     ...: b = np.array([150, 154, 147, 126, 148, 125])
     ...: 

In [657]: mask = a!="b"

In [658]: mask
Out[658]: array([ True, False,  True,  True, False,  True], dtype=bool)

# Crux of the implmentation happens here :
In [696]: np.where(mask,np.arange(mask.size),0)
Out[696]: array([0, 0, 2, 3, 0, 5])

In [697]: np.maximum.accumulate(np.where(mask,np.arange(mask.size),0))
Out[697]: array([0, 0, 2, 3, 3, 5])# Stepped indices "intervaled" at masked places

In [698]: idx = np.maximum.accumulate(np.where(mask,np.arange(mask.size),0))

In [699]: b[idx]
Out[699]: array([150, 150, 147, 126, 126, 125])

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